$k$ is some positive number which can be chosen freely for example $k=1$ or $k=3/2$ or $k=2/3$. $f_0$ is only a function of $\lambda_0$, likewise $f_1$ is only a function of $\lambda_1$.

If I would insert $\lambda_0$ from $f_0$ into $f_2$ and insert $\lambda_1$ from $f_1$ into $f_2$ then $f_2$ will be a function of only $x,k,y_0,y_1$.

In this case I would like to have a 3D plot with x axis =$y_0$, y-axis=$y_1$, and z-axis =$x$ for $7/81<y_0<1/9$ and $7/81<y_1<1/9$, for some known $k$ as mentioned above and $x$ must be eventually in $0<x<1$, by positivity condition. Is it doable? how can I do that?

EDIT:

I obtain the polynomials via equating them to $0$. Namely, $f_0=0$, $f_1=0$ and $f_2=0$ are all given and known.

@Algohi that should be a 3D function. Let me explain. We can write it as $x=f(y_0,y_1)$ and for each $(y_0,y_1)$ we have a value of $x$. I am only interested in the values of $x$ which are in $[0,1]$. OK I think I forgot to mention that I have $f_0=0$, $f_1=0$, $f_2=0$! I must edit. Sorry for that.
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Seyhmus GüngörenJul 12 '14 at 19:31

see at last step, I specify one of the values of $\lambda_0$ and $\lambda_1$, i.e. the first value. you can choose any combination like (1,1),(1,2),(2,1)or(2,2).

Then comes the second difficulty. You may not get a graphics output with ContourPlot3D because the final function returns a lot of complex values. You can refine your function or see the Abs, Re or Im value.

Based on the answer of Sumit I manipulated the code so that the code should make what I want. But NSolve takes alot of time. I am still waiting and I dont think that the code will end. Is there a faster way? because I am interested in only the 3D plot which can be obtained purely numerically..

I think the problem is with the complicated initial functions, which makes the final solution quite difficult, both for Solve and NSolve. That's why I suggest to go with Abs, Re or Im. Your method looks quite good to me. As a cross check generate Table[{x,y0,y1,f[x,y0,y1]},{x,..},{y0,..},{y1,..}]//TableForm and you can get an idea about if there exist a solution in the defined range.
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SumitJul 13 '14 at 9:23

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